I am solving Linear Algebra and stuck in these kind of model problems.
Let $V$ be the set of all real valued functions $y = f(x)$ satisfying $y'''-7y'-6y=0$. Show that $V (\Bbb {R})$ is a 3-dimensional real vector space. Write Basis of this vector space.
How to solve the above Differential Equation and proceed further?
Best Answer
This is a linear differential equation with constant coefficients.
If we assume $y=e^{ax}$, we get $(a^3-7a-6)e^{ax}=0$.
Since $e^{ax}\ne0$, that means $\color{blue}{a^3-7a-6=0}$.
Say the solutions to that $\color{blue}{\text{cubic equation}}$ are $a_1, a_2$, and $a_3$.
Then $y=e^{a_1x}, e^{a_2x}$, and $e^{a_3x}$ are solutions to the differential equation.
Since this is a homogenous differential equation,
any linear combination of solutions will also be a solution.
The general solution is thus $y=c_1e^{a_1x}+c_2e^{a_2x}+c_3e^{a_3x}$.